What is the first derivative of r with respect to t (i.e., differentiate r with respect to t)? r = 5/(t2)Note: Use ^ to show exponents in your answer, so for example x2 = x^2. Also, type your equation answer without additional spaces.

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Answer 1

Answer:

Answer:

The first derivative of $r(t) = 5\cdot t^(-2)$ (r(t)=5*t^{-2}) with respect to t is $r'(t) = -10\cdot t^(-3)$ (r'(t) = -10*t^{-3}).

Step-by-step explanation:

Let be $r(t) = (5)/(t^(2))$ , which can be rewritten as $r(t) = 5\cdot t^(-2)$ . The rule of differentiation for a potential function multiplied by a constant is:

$(d)/(dt)(c \cdot t^(n)) = n\cdot c \cdot t^(n-1)$ , $\forall \,n\neq 0$

Then,

$r'(t) = (-2)\cdot 5\cdot t^(-3)$

$r'(t) = -10\cdot t^(-3)$ (r'(t) = -10*t^{-3})

The first derivative of $r(t) = 5\cdot t^(-2)$ (r(t)=5*t^{-2}) with respect to t is $r'(t) = -10\cdot t^(-3)$ (r'(t) = -10*t^{-3}).

-6

Step-by-step explanation:

The given translation is

$T_(-3,-8) (x,y)$

Which means that the original figure will be moved 3 units to the right and 8 units downwards.

Remember, when it comes to translations, when we subtract units to $x$ that means the figure will be moved rightwards. And if we subtract units from $y$ , that means the figure will be moved downwards.

So, the original figure has as vertex B(1,2). If we apply the transformation to its vertical cordinate $y=2$ , we would have

$y'=2-8=-6$

Therefore, the right answer is the third choice -6.

Question 12 of 22What is sin 60°?
O A.
OB.
O C. 1
O D.
2
1
-
√2
OE 1/1/20
OF. √3

What is the first derivative of r with respect to t (i.e., differentiate r with respect to t)? r = 5/(t2)Note: Use ^ to show exponents in your answer, so for example x2 = x^2. Also, type your equation answer without additional spaces.

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